Let $G$ be a group and $CG$ the complex group algebra over the field $C$ of complex number. The group von Neumann algebra $NG$ is the completion of $CG$ wrt weak operator norm in $B(l^2(G))$, the set of all bounded linear operators on Hilbert space $l^2(G)$. Let $f:G \to H$ be any homomorphism of groups. My question is: is there a homomorphism of the group von Neumann algebra $NG \to NH$ induced from $f$?.

If $NG$ is replaced with $CG$, it's obvious true. If $NG$ is replaced with $C^\ast_r(G)$, the reduced group $C^\ast$ algebra, it's not necessary true.

3 Answers
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By the way, here's the "correct" functorial property. If G and H are abelian, and $f:G\rightarrow H$ is a continuous group homomorphism, then we get a continuous group homomorphism $\hat f:\hat H\rightarrow \hat G$ between the dual groups. By the pull-back, we get a *-homomorphism $\hat f_*:C_0(\hat G) \rightarrow C^b(\hat H)$. We should think of $C^b(\hat H)$ as the multiplier algebra of $C_0(\hat G)$. Then $C_0(\hat G) \cong C^*_r(G)$, and so we do get a *-homomorphism $C^*_r(G) \rightarrow M(C^*_r(H))$; the strict-continuity extension of this is a *-homomorphism $M(C^*_r(G)) \rightarrow M(C^*_r(H))$ which does indeed send $\lambda(s)$ to $\lambda(f(s))$.

For non-abelian group (in fact, non-amenable groups) it's necessary to work with $C^*(G)$ instead.

We cannot ensure a map to $C^*_r(H)$ itself, as we cannot ensure a map from $C_0(\hat G)$ to $C_0(\hat H)$; indeed, this would only happen when $\hat f$ were a proper map.

Similarly, we don't get maps at the von Neumann algebra level, as we don't get a map $L^\infty(\hat G) \rightarrow L^\infty(\hat H)$: we would need that $\hat f$ pulled-back null sets in $\hat G$ to null sets in $\hat H$.

The homomorphism $f$ extends to a homomorphism of reduced group $C^{\ast}$-algebras if and only of $\ker(f)$ is amenable, and extends to a homomorphism of group von Neumann algebras if and only if $\ker(f)$ is finite.

Nice to know, but how to use amenability or finitness of \kerf? Can you give some hints or references?
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m07klOct 23 '10 at 14:35

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If $\ker(f)$ is finite, then the normalized sum over all elements in $\ker(f)$ is a central projection $p \in N(G)$ and the extension is given by the cut-down composed with the inclusion of $G/\ker(f) \subset H$, i.e. $N(G) \to pN(G)p = N(G/\ker(f)) \subset N(H)$.
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Andreas ThomOct 23 '10 at 16:03

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If $\ker(f)$ is amenable, then take a sequence of Foelner sets $F_n \subset \ker(f)$ and define vectors $$\xi_n = |F_n|^{-1} \cdot \sum_{g \in F_n} \delta_g.$$ Then, the states $\phi_n(a) = \langle a \xi_n, \xi_n \rangle$ on $C^{\ast}_{red}(G)$ will have a weak limit and the associated GNS-representation gives a $\ast$-homomorphism $C^{\ast}_{red}(G) \to C^{\ast}_{red}(G/\ker(f))$. Again, this can be composed with the inclusion $C^{\ast}_{red}(G/\ker(f)) \subset C^{\ast}_{red}(H).
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Andreas ThomOct 23 '10 at 16:09

Well, for $s\in G$ let $\lambda(s)$ be the left-translation operator by $s$; all such operators are in the group von Neumann algebra. I guess that the hoped for homomorphism $F:NG \rightarrow NH$ should satisfy $F(\lambda(s)) = \lambda(f(s))$ for $s\in G$, and we should have that $F$ is an (ultraweakly) continuous $*$-homomorphism. In particular, $F$ is contractive.

Then $F$ need not exist. Let $G=\mathbb Z$ and $H=\mathbb Z/n\mathbb Z$. Then $NH = CH$, so we have a trace on $NH$ which sends $\lambda(0)$ to $1$. So if we apply $F$, and then take the trace, we should get an ultraweakly continuous functional $\phi$ on $NG$ which satisfies $\phi(\lambda(ns)) = 1$ for all $s\in\mathbb Z$.

But this can't happen: maybe we can see this via the Fourier transform. Then $NG \cong L^\infty(\mathbb T)$ and $\phi$ induces $h\in L^1(\mathbb T)$ which satisfies $\int h(\theta) e^{ins\theta} d\theta = 1$ for all $s\in\mathbb Z$. This violated Reimann-Lebesgue.

On the other hand, if $G \subseteq H$ is an inclusion (of discrete groups, to avoid topology) then we do get an inclusion $NG \rightarrow NH$. Here's a construction. Find an index set $I$ and $(h_i)$ in $H$ such that $H$ is the disjoint union of $\{Gh_i\}$. Then define $V:l^2(H)\rightarrow l^2(G)\otimes l^2(I)$ by $V(\delta_h) = \delta_g\otimes\delta_i$ if $h=gh_i$ (so defined on point-masses, and extend by linearity). So $V$ is unitary, and $\theta:x\mapsto V^*(x\otimes 1)V$ is a normal $*$-homomorphism $NG\rightarrow B(l^2(H))$. Then, for $r\in G$, $V^*(\lambda(r)\otimes 1)V(\delta_h) = V^*(\delta_{rg}\otimes\delta_i) = \delta_{rh}$ as $rg\in G$. So $\theta$ maps into $NH$, and does what we want.

Is this somehow connected to the fact that, since the inverse image of an element of $H$ may be infinite, the map $G\to H$ does not induce a map $\ell^2(H)\to\ell^2(G)$?
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Harald Hanche-OlsenFeb 11 '10 at 16:39

Matt: what was wrong with the statement you struck through? If G is a closed subgroup of H then the sub-vN-algebra of N(H) that is generated by "left translation by elements of G" is isomorphic to N(G), is it not? I thought this was in Takesaki & Tatsuuma, for instance.
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Yemon ChoiFeb 11 '10 at 18:41

Takesaki & Tatsuuma certainly study how one can "see" closed subgroups of G as certain subalgebras of NG, but they seem to studiously avoid claiming that the subalgebras are isomorphic to NH, at least on my reading. I struck through the sentence, as the proof I had was (obviously) wrong.
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Matthew DawsFeb 11 '10 at 21:11

Just to note (mostly for amusement) that in Matt's counterexample, one could take $n=1$. (Thus: the augmentation character is well-defined as a functional on the $\ell^1$-group algebra, here being evaluation of the Fourier transform at $1$; but it doesn't extend in the right way to VN(Z), because one can't evaluate an element of $L^\infty(T)$ at the point $1$.
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Yemon ChoiFeb 12 '10 at 6:26